1. List 转 Map<Integer,List>
Map<Integer, List<User>> subLineInfoMap = userInfos .stream().collect(Collectors.groupingBy(User::getSex));
2. List 转 Map<Integer,User>
Map<Long, User>userMap = userInfos.stream().collect(Collectors.toMap(User::getStuNum,Function.identity()));
3. List 转 Map<Integer,String>
Map<Integer, String> collect = userInfos.stream().collect(Collectors.toMap(User::geUserNum, User::getUserName));
4. list 转 map 保持顺序
LinkedHashMap<String, User> userMap = users.stream().collect(LinkedHashMap::new, (map, item) -> map.put(item.getAccountId(), item), Map::putAll);
5. 将list转成map 并排序
将list 排序，并按照排序后的结果进行有序分组
LinkedHashMap<String, List<AlarmData>> alarmMap = alarmDataList.stream().sorted(Comparator.comparing(t->t.getId().getData_time())).collect(Collectors.groupingBy(t->t.getId().getVirtualPointId(), LinkedHashMap::new, Collectors.toList()));
将map排序，并且每个key对应的list里面也是排序好的
6. 我们在利用Lambda 将list转成Map时就会出现 Duplicate key xxxx 的异常,意思就是对要转为map的key有重复了,除了进行for循环去重之外,我们还有其它方式能够优雅的处理它.
key重复时直接用后面的值(使用最新的或最老的值)
Map<String, Long> collect = list.stream().collect(Collectors.toMap(User::getExternalUserId, User::getUserId, (val1, val2) -> val2));
将两个值拼接起来
Map<String, Long> collect = list.stream().collect(Collectors.toMap(User::getExternalUserId, User::getUserId, (val1, val2) -> val1+val2));
将重复key的value变成一个集合，注意null值处理
Map<String, List<String>> tagMap = list.stream().collect(Collectors.toMap(User::getExternalUserId, s -> { List<String> tags = new ArrayList<>(); tags.add(s.getFollowTags()); return tags; }, (List<String> val1, List<String> val2) -> { val1.addAll(val2); return val1; } ));